Question

我正在尝试创建一个递归的Python函数，它接受一个句点列表并将它们合并到一个干净的时间轴中。它应该扫描列表并应用这些规则：

如果在期间内找到无的值：将date替换为datetime.date.today（）
如果句点在内开始，在另一段时间内结束：将其删除。
如果句点在之前开始，但在的另一段时间内结束：延长开始日期。
如果句点在之后结束，但在另一段时间内开始：延长结束日期。
如果一段时间在之后开始而在之后的另一段时间结束：保留它，它是一个单独的时段。 < / p>
如果一段时间在之前开始而在之前结束另一段时间：保留它，它是一个单独的时段。 < / p>

给出一个输入和所需输出的例子可能要容易得多（假设值是用datetime格式化的）：

[I] = [(01/2011, 02/2015), (04/2012, 08/2014), (09/2014, 03/2015), (05/2015, 06/2016)]
[O] = [(01/2011, 03/2015), (05/2015, 06/2016)]  
# Notice how the output has produced a set of minimum length whilst covering all periods.

[I] = [(07/2011, 02/2015), (04/2012, 08/2014), (09/2014, 04/2015), (06/2015, None)]
[O] = [(07/2011, 04/2015), (06/2015, date.today())]
# Also, notice how the output has amended None, so it can compare dates.

感谢@khredos，我写了以下内容，但它仍然没有输出所需的最小字符串：

from datetime import datetime

# Here is an example list of time periods
periods = [('01/2011', '02/2015'), ('04/2012', '08/2014'), ('09/2014', '03/2015'), ('05/2015', '06/2016')]

# this lambda function converts a string of the format you have specified to a 
# datetime object. If the string is None or empty, it uses today's date
cvt = lambda ds: datetime.strptime(ds, '%m/%Y') if ds else datetime.today()

# Now convert your original list to an iterator that contains datetime objects
periods = list(map(lambda s_e : (cvt(s_e[0]), cvt(s_e[1])), periods))

# Next get the start dates into one list and the end dates into another
starts, ends = zip(*periods)

# Finally get the timeline by sorting the two lists
timeline = sorted(starts + ends)

# Output: [datetime.datetime(2011, 1, 1, 0, 0), datetime.datetime(2012, 4, 1, 0, 0), datetime.datetime(2014, 8, 1, 0, 0), datetime.datetime(2014, 9, 1, 0, 0), datetime.datetime(2015, 2, 1, 0, 0), datetime.datetime(2015, 3, 1, 0, 0), datetime.datetime(2015, 5, 1, 0, 0), datetime.datetime(2016, 6, 1, 0, 0)]

Answer 1

from datetime import datetime

# Here is an example list of time periods
periods = [('01/2011', '02/2015'), ('04/2012', '08/2014'), ('09/2014', '03/2015'), ('05/2015', '06/2016')]

# this lambda function converts a string of the format you have specified to a 
# datetime object. If the string is None or empty, it uses today's date
cvt = lambda ds: datetime.strptime(ds, '%m/%Y') if ds else datetime.today()

可用格式为here

# Now convert your original list to an iterator that contains datetime objects
periods = list(map(lambda s_e : (cvt(s_e[0]), cvt(s_e[1])), periods))

# Next get the start dates into one list and the end dates into another
starts, ends = zip(*periods)

# Finally get the timeline by sorting the two lists
timeline = sorted(starts + ends)

输出应类似于

[datetime.datetime(2011, 1, 1, 0, 0), datetime.datetime(2012, 4, 1, 0, 0), datetime.datetime(2014, 8, 1, 0, 0), datetime.datetime(2014, 9, 1, 0, 0), datetime.datetime(2015, 2, 1, 0, 0), datetime.datetime(2015, 3, 1, 0, 0), datetime.datetime(2015, 5, 1, 0, 0), datetime.datetime(2016, 6, 1, 0, 0)]

尝试使用任何日期列表，您应该观察到相同的行为。

HTH

递归列表函数

1 个答案: